The equation of a curve is 3x² + 4xy + 3y² = 5.
- (a)Show that (dy/dx) = -(3x+2y)/(2x+3y)[4]
- (b)Hence find the exact coordinates of the two points on the curve at which the tangent is parallel to y + 2x = 0.[5]
Mark scheme
Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.
- B1 State or imply 6x as the derivative of 3x² or imply (dy/dx) as the derivative of 3y² [Allow y' for dy/dx throughout. Accept ∂f/∂x = 6x + 4y .]
- B1 State or imply 4x(dy/dx) + 4y as the derivative of 4xy [Accept ∂f/∂y = 4x+6y.]
- M1 Equate derivative of LHS to zero and solve for (dy/dx) [Allow an extra dy/dx in front of their differentiated equation. Allow if '= 0' is implied but not seen. Allow dy/dx = -(∂f/∂x)/(∂f/∂y)]
- A1 Obtain (dy/dx) = -(3x+2y)/(2x+3y) [AG - must come from correct working. The position of the negative must be clear.]
- M1 Equate (dy/dx) to -2 and solve for x in terms of y or for y in terms of x [Must be using the given derivative.]
- A1 Obtain x = -4y or y = -x/4 [Seen or implied by correct later work.]
- DM1 Substitute their x = -4y or their y = -x/4 in curve equation [Allow unsimplified.]
- A1 Obtain y = ±(1/√7) or x = ±(4/√7) [Or exact equivalent.]
- A1 Obtain both pairs of values [Or x = -(4/√7) and y = 1/√7 or x = (4/√7) and y = -(1/√7) or exact equivalent. A1 A0 for incorrect final pairing.]
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